7.3.1. Do two processes have the same mean?

from two independent processes (the \(Y\) values are sampled from process 1 and the \(Z\) values are sampled from process 2), there are three types of questions
regarding the true means of the processes that are often asked. They are:

Are the means from the two processes the same?

Is the mean of process 1 less than or equal to the mean of process 2?

Is the mean of process 1 greater than or equal to the mean of process 2?

Form of the test statistic where the two processes do NOT have
equivalent standard deviations

If it cannot be assumed that the standard deviations from the two processes
are equivalent, the test statistic is
$$ t = \frac{\bar{Y} - \bar{Z}}
{\sqrt{\frac{s_1^2}{N_1} + \frac{s_2^2}{N_2}}} \, . $$
The degrees of freedom are not known exactly but can be estimated using
the Welch-Satterthwaite approximation
$$ \nu = \frac{\left( \frac{s_1^2}{N_1} + \frac{s_2^2}{N_2} \right)^2}
{\frac{s_1^4}{N_1^2(N_1-1)} + \frac{s_2^4}{N_2^2(N_2-1)}} \, . $$

Test strategies

The strategy for testing the hypotheses
under (1), (2) or (3) above is to calculate the appropriate t
statistic from one of the formulas above, and then perform a test at
significance level α, where α is chosen to be small, typically .01, .05 or .10. The hypothesis associated with each case enumerated above
is rejected if:

\(|t| \ge t_{1-\alpha/2, \, \nu}\)

\(t \ge t_{1-\alpha, \, \nu}\)

\(t \le t_{\alpha, \, \nu}\)

Explanation of critical values

The critical values from the \(t\)
table depend on the significance level
and the degrees of freedom in the standard deviation. For hypothesis (1),
\(t_{1-\alpha/2, \, \nu}\)
is the \(1-\alpha/2\)
critical value from
the t table with \(\nu\)
degrees of freedom and similarly for hypotheses (2) and (3).

For a one-sided test at the 5 % significance level, go to the
t table for 0.95 signficance
level, and look up the critical value for degrees of freedom \(\nu\)
= 16. The critical value is 1.746. Thus, hypothesis (2) is rejected
because the test statistic (\(t\)
= 2.269) is greater than 1.746 and, therefore, we conclude that
process 2 has improved assembly time (smaller mean) over process 1.